Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 102


$sec~(cos^{-1}~\frac{u}{\sqrt{u^2+5}}) = \frac{\sqrt{u^2+5}}{u}$

Work Step by Step

Let $~~\theta = cos^{-1}~\frac{u}{\sqrt{u^2+5}}$ Then $~~cos~\theta = \frac{u}{\sqrt{u^2+5}}$ We can find an expression for $sec~\theta$: $sec~\theta = \frac{1}{cos~\theta}$ $sec~\theta = \frac{\sqrt{u^2+5}}{u}$ Therefore, $~~sec~(cos^{-1}~\frac{u}{\sqrt{u^2+5}}) = \frac{\sqrt{u^2+5}}{u}$
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